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2 years ago

Y=x-3 y=2 equations graph

Mathematics

1 answer:

katovenus [111]2 years ago

5 0

Answer:

x=5

Step-by-step explanation:

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Molly has a jar containing 4 yellow, 3 blue, 3 green, and 2 orange marbles

NeTakaya

Answer: .111 with a bar notation over the last 1 or 11.1%

Step-by-step explanation:

There are 12 marbles in total. There are 4 yellow marbles. Therefore there is a 4/12 probability of Randomly picking a Yellow marble the first time . Since she placed the marble back, there is a 4/12 probability of getting a yellow marble the 2nd time.

(4/12)(4/12) = 16/144 = 1/9 = .111 with a bar notation over the last 1 or 11.1% (about 11%)

6 0

3 years ago

Can one of you lovely people help me please ?

saveliy_v [14]

Answer:

the answer is

C. = x=22.82

4 0

2 years ago

Read 2 more answers

3. What is the scale factor of Figure B to Figure A?

xxMikexx [17]

Answer:

2.5

Step-by-step explanation:

From the diagram, figure B was enlarged to obtain figure A.

The two figures are therefore similar.

The corresponding sides are in the same proportion. That constant value of the proportion is called scale factor.

It is given by:

$k = \frac{image \: length}{corresponding\:object \: length}$

Figure B is the image of A

$k = \frac{10}{4} = \frac{25}{10} = \frac{21.5}{8.6} = 2.5$

Therefore the scale factor is 2.5

4 0

3 years ago

Isla flipped a coin 30 times. The coin landed heads up 9 times and tails up 21 times.

DedPeter [7]

A.) Either way the probability would be 1/2 because there are only 2 sides of a coin so fate decides on which side it will land on.

B.)Same as A

8 0

3 years ago

Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

8 0

2 years ago